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LED question

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  • LED question

    I have a laser etched acrylic block that I have made a wooden case for. The block is 3" x 3" x 6" tall. I have routed a 3 x 3 recess in wooded blocks to hold the acrylic block tight. I would like to put an led in the wooden top to shine directly into the block. It will be a tight squeeze. I'm concerned about the heat the led will give off.

    Here is the led I'm thinking about.

    How much heat does an led give off?

    I have considered making a 1/4" thick copper strip to fasten the star to. Making it long enough to go out the back side of the wooden case to draw off the heat. Would that be sufficient?

  • #2
    How much heat depends on what power level (milliamps) you drive it. 20mA should require no cooling and should be bright enough in my opinion. More power than that, and "it depends".


    • #3
      If you feed it nominal current (350 mA) it will create about .5 watts. What that translates to in temperature depends on the mechanical coupling between the lamp and the block or what ever the lamp is attached to. Boltzmann's constant plus some material unknowns means try it. It may work just fine.


      • #4
        Operating voltage for that led is 3 volts at 350 ma. That's just over 1 watt. Yes, I think a copper strip leading outwards would be a good idea, and it should expand into about 3 sq inches of air contact area. Put it another way, a piece of copper about 1 x 3 inches, with a tab leading off to where the led would mount would do it.

        That's driving it pretty significantly. I agree, it would be plenty bright at a fraction of that current. I would still mount it to some kind of heat spreader though- typically it would be on a pc board, with enough copper area to dissipate the heat. If you ran it at say 100 ma, a piece of pc board large enough to make connections to would probably be enough. The wires themselves will conduct the heat away.
        Last edited by darryl; 01-15-2014, 12:12 AM.
        I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-


        • #5
          The general formula for the heat generated in an LED is simply:

          P = V x I

          Power is the heat generated in Watts,

          V is the operating Voltage of the LED in Volts, and

          I is the current that it is operating at in Amps.

          The operating Voltage is given on the spec sheet or you can just measure across the terminals while it is operating. Generally you set up a series resistor to determine the operating current. The formula here is

          R = (Vs-V)/I

          R is the resistance in Ohms

          Vs is the supply Voltage in Volts,

          V is the operating Voltage of the LED in Volts (same as above), and

          I is the operating current that you want in Amps.

          So if you have an LED that operates at 2 V and you have a 5 Volt supply and want to operate it with 25 mA of current:

          R = (5V-2V)/0.025A
          R = 3V/0.025A
          R = 120 Ohms

          and the power in the LED is:

          P = V x I
          P = 2V x 0.025A
          P = 0.05 Watts

          Now the power dissipated in the resistor is calculated separately, using the same formula, but the Voltage drop across the resistor instead of that which is across the LED.

          Pr = V x I
          Pr = 3V x 0.025A
          Pr = 0.075 Watts

          If the LED package contains the current limiting resistor then it is all combined in that one package and the same power formula applies. Using the same values as above, but with the resistor and LED in one package:

          P = V x I
          P = 5V x 0.025A
          P = 0.125 Watts

          5 Volts is the supply Voltage applied to the LED/resistor package. All of the power is now dissipated in that one package.

          In all of the above power calculations I have neglected the amount of power that is converted to light and that leaves the LED to be converted to heat elsewhere. However, even LEDs are relatively inefficient at converting electric power to light and the actual fraction that becomes heat will be less than the total power that is present in the electricity consumed so you can safely neglect this factor. You will just have a bit of a safety margin.
          Paul A.
          SE Texas

          And if you look REAL close at an analog signal,
          You will find that it has discrete steps.


          • #6
            Hi Tim
            How much heat does an led give off?
            The real question is what will the junction temperature of the LED be. The answer is:
            Junction Temp=Ambient Temp + (Volts * Amps * Junction to air thermal resistance measured in degrees C/Watt)
            The datasheet gives a thermal resistance of 10C/W junction to pad for the LED. You need to add to that the thermal resistances
            of solder joint to star + star to heatsink + heatsink to ambient.
            This site should allow you to get a good approximation of the heatsinking requirements:
            Google turned up that 1 square inch of 1/2 ounce copper has a thermal to air resistance of about 50C/W. To lower that, you have to
            use thicker copper in addition to increasing surface area. Vertical surfaces are preferred to promote convection.
            Location: Long Island, N.Y.


            • #7
              not to be a pain, but red LEDs are usually around 2V compared with white ones at ~3V. One of those at 350mA will be very bright, just depends on what you want to use it for. A rough rule of thumb for heatsinking is 10in.sq. per watt in still air, ~2in.sq/W in moving air, so at 350mA you're looking at ~700mW or about of surface area. THat's pretty conservative though as it depends on how long it's going to be running for, ambient temp, your tolerance to very hot LEDs etc.

              How are you going to drive it? Do you have a driver in mind?


              • #8
                You don't need near that much to do the job.

                I do this with 4 small LED's running 20ma at around 2 volts, and 3 AAA batteries. The LED's are good for around 600 mcd each, battery life runs around 30 hours. Keep in mind that the deeper the etching and the darker the ambient lighting, the more it will stand out.
                Definition: Racecar - a device that turns money into noise.


                • #9
                  Originally posted by mattthemuppet View Post
                  One of those at 350mA will be very bright,
                  ^^^ This!

                  350mA is great... for a flashlight illuminating an entire room. 20mA from a modern high-flux LED like the Cree XP-E is going to be plenty to illuminate a 3x3x6 acrylic cube... at 350mA you will want sunglasses!

                  5mm LEDs can run at 20mA all year long with no more heatsinking than that provided by their two skinny steel leads, which ain't much. The XP-E SMT emitter is even more robust, so I don't see a problem at 20mA and no heatsinking besides whatever it is mounted to.


                  • #10
                    Well originally I planned to drive it at 350mA. As a test I shined my "Larry" led flashlight down thru the top of the cube, it is 60 lumen per the manufacturer. I like that level of light. The Cree XP-E in red-orange according to the data sheet states 57 lumens. So now I'm not sure what I should drive it with. I have not found a dimmable driver that would go as low as 20mA. Any suggestions.

                    Thanks for all the help.


                    • #11
                      Does it need to be dimmable, or just run at 20mA? If you just want to run it at a fixed current, a series resistor from a power supply will be fine. Or this circuit with a 78L05 regulator will supply the LED with a constant 20mA of current.
                      I show a 6V battery as an example power source, but the circuit will be fine up to about 15 volts or so input. I used to sell these prewired a while back.
                      Last edited by lwalker; 01-15-2014, 07:11 PM.


                      • #12
                        What did you have in mind as a power source? A wall wart? Just pick one that is 5 or 6 volt and use a series resistor- value would be anywhere from about 15 to 150 ohms. It doesn't sound like you want it dimmable, but just to get a brightness level that looks good. No driver required.
                        I seldom do anything within the scope of logical reason and calculated cost/benefit, etc- I'm following my passion-


                        • #13
                          It doesn't necessarily need to be dimmable. I'm just not sure what light level I will want.

                          I thinking wall wart. I could try a few different resistors as a test.